4.38: Electrophoresis

A laboratory is setting up an electrophoretic method for identification of serum proteins; however, the appearance of the protein bands appears to be too diffuse. The technologist notes that the pH of the freshly prepared buffer is correct but a bit warm; the apparatus appears to be warm to the touch as well. One possibility is that the problem could be due to excess heating (Table 10-5, p. 213). The amount of heat generated by an electric current is related to the power (wattage) dissipated by the system. The relationship between wattage, current (i), and resistance (R) is: $$w = i^{2}R$$The technologist has a meter which can read both voltage and amperage. During an electrophoretic run, these values are recorded as 100 volts and 0.01 amps. The power generated by the assay, according to the manufacturer, is not to exceed 0.5 watts.

QUESTIONS

1. Does an excess amount of power appear to be the problem?
2. After discussion with the clinical chemist, it is decided to reduce the power passing through the system. What approaches are available to do this?
3. After much discussion, it is decided that the length of time needed for the assay should be as short as possible in order to complete the assay conveniently in a working day. With this additional information, what is the best change to make in the assay system to correct the original problem?
4. The buffer solution in this assay contains 0.1 molar sodium barbital (Na Bar) and 0.01 molar magnesium phosphate (MgPO₄). How much would this buffer need to de diluted to reduce the power to 0.5 watts? What will be the ionic strength of the diluted buffer?

Questions to Consider

1. What is the relationship between voltage, amperage, and resistance?
2. What is the power at 100 volts and 0.01 amps?
3. What is the relationship between voltage, current, and power?
4. What is the relationship between ionic strength of an electrophoresis buffer, the current passing through it, and the temperature generated in the buffer?
5. How does changing the voltage affect the rate of migration (velocity) of molecules in an electrophoretic field?
6. What is ionic strength of a solution calculated? What is the ionic strength of the current buffer?

Answer

1. This amount of power, 1.0 watt, exceeds the manufacturer’s specifications.
2. In order to decrease the power, either i or R must be decreased. From equation 10.3, if R is decreased at constant voltage, i will increase proportionately. However, power increases as the square of the value of 1 and only proportionately to R. Therefore, if R is decreased, there will be a net increase in power. Thus, to decrease power at constant voltage, R must be increased, causing a decrease in i. Alternatively, if R is held constant, decreasing the voltage will decrease i and thus reduce the power. If voltage is to remain constant, then R must be increased.
3. Simply decreasing voltage to reduce the power is not an acceptable solution in this particular case. The best solution is to increase resistance by decreasing ionic strength.
4. By diluting this buffer two-fold, the ionic strength will decrease by one half as will the conductivity, increasing the resistance by two-fold and decreasing current by one half (at a constant voltage): $$\begin{split} R &= 2 \times 10,000\; ohms = 20,000\; ohms \\ i &= \frac{1}{2} \times 0.01\; amps=0.005\; amps \end{split}$$The wattage of this system is now. $$\begin{split} W &= i^{2} R \\ &= (0.005)^{2} \times 20,000 = 0.5\; watts \end{split}$$This will decrease the heat produced by the system and increase the sharpness of the bands, but time for analysis will remain the same.

Answers to Questions to Consider

1. The relationship between voltage, amperage, and resistance is: $$V = R \times i$$(Equation 10-3, p. 204).
2. In order to calculate wattage from the information given, the resistance must first be determined. Using Equation 10-3, $$\begin{split} V &= Ri \\ R &= \frac{V}{i} \\ &= \frac{100\; volts}{0.01\; amps} \\ &= 10,000\; ohms \end{split}$$Now, using the relationship between power, current, and resistance, W can be calculated as follows: $$\begin{split} W &= i^{2} R \\ &= (0.01\; amps)^{2} (10,000\; ohms) \\ &= 1.0\; watt \end{split}$$
3. By combining the equations in Answer 2, it is seen that W = iV.
4. Ionic strength is proportional to current; the greater the ionic strength, the greater the current the solution is capable of passing at the same voltage. Increased current increases wattage and therefore results in more heat generated by the system. See p 210-211.
5. The rate of migration is directly related to the voltage. The time the electric field is allowed to act on the molecule is directly related to resolution. Thus, decreasing voltage will result in a proportional decrease in rate of migration and increase in the amount of time required to achieve the same degree of resolution, i.e. for all the molecules to travel the same distance.
6. The ionic strength (I) is related to the concentration of the ions and their charge by the following equation: $$I = \frac{1}{2} C_{1} Q_{1}^{2} + \frac{1}{2} C_{2} Q_{2}^{2} + \ldots \frac{1}{2} C_{N} Q_{N}^{2}$$In this case $$\begin{split} I &= \frac{1}{2} C_{Na} Q_{Na}^{2} + \frac{1}{2} C_{Bar} Q_{Bar}^{2} + \frac{1}{2} C_{Mg} Q_{Mg}^{2} + \frac{1}{2} C_{PO_{4}} Q_{PO_{4}}^{2} \\ &= \frac{1}{2} (0.1)(1)^{2} + \frac{1}{2} (0.1)(1)^{2} + \frac{1}{2} (0.01)(2)^{2} + \frac{1}{2} (0.01)(2)^{2} \\ &= 0.14\; molar \end{split}$$